Question

The boiling point of ethanol is 78.4 °C, and the enthalpy change for the conversion of liquid to vapor is ΔHvap = 38.56 kJ/mol. What is the entropy change for vaporization, ΔSvap, in J/(K ⋅ mol)?

Answer 1

Step-by-step explanation:

The given data is as follows.

Enthalpy of vaporization,
`\Delta H_(vap)` = 38.56 kJ/mol

Temperature (T) =
`78.4^(o)C`

= (78.4 + 273) K

= 351.4 K

Now, we will calculate the change in entropy using the formula as follows.

`\Delta H_(vap) = T \Delta S_(vap)`

`\Delta S_(vap) = (\Delta H_(vap))/(T)`

=
`(38.56 kJ/mol)/(351.4 K)`

= 0.109 kJ/mol K

Thus, w ecan conclude that the entropy change for vaporization is 0.109 kJ/mol K.